LG P6156 简单题

( ext{Problem})

( ext{Analysis})

显然 (f=mu^2)
那么

[egin{aligned} sum_{i=1}^n sum_{j=1}^n (i+j)^k &= sum_{d=1}^n mu^2(d) d^{k+1} sum_{i=1}^{lfloor frac{n}{d} floor} sum_{j=1}^{lfloor frac{n}{d} floor} (i+j)^k [gcd(i,j)=1] \ &= sum_{d=1}^n mu^2(d) d^{k+1} sum_{g=1}^{lfloor frac{n}{d} floor} mu(g) g^k sum_{i=1}^{lfloor frac{n}{dg} floor} sum_{j=1}^{lfloor frac{n}{dg} floor} (i+j)^k \ end{aligned} ]

我们考虑预处理

[f_1 = sum_{i=1}^n mu^2(d) d^{k+1} \ f_2 = sum_{i=1}^n mu(d) d^k \ f_3 = sum_{i=1}^n sum_{j=1}^n (i+j)^k ]

这样就可以数论分快套数论分快搞定

那么就考虑如何预处理这三个前缀和
显然 (g(d)=d^k) 是个积性函数，于是可以线筛处理处所有 (d^k)
那 (f_1) 和 (f_2) 一遍就出来了

现在就看 (f_3) 了
我们对 (f_3) 差分

[egin{aligned} f_3(n)-f_3(n-1) &= sum_{i=1}^n sum_{j=1}^n (i+j)^k - sum_{i=1}^{n-1} sum_{j=1}^{n-1} (i+j)^k \ &= 2 sum_{i=1}^n (n+i)^k - (2n)^{k} end{aligned} ]

也就是说我们处理出 (sum_{i=1}^{2n} d^k) 就可以处理出这个 (f_3) 的差分数组
然后再做一遍前缀和就可以得到 (f_3)

到此本题就结束了
注意空间！！

( ext{Code})

#include<cstdio>
#include<iostream>
#define re register
using namespace std;
typedef long long LL;

const int N = 1e7, P = 998244353;
LL k;
int totp, n;
int pr[N], vis[N + 5], mu[N + 5], pk[N + 5], spk[N + 5], f1[N / 2 + 5], f2[N / 2 + 5], f3[N / 2 + 5];

inline int fpow(LL x, LL y)
{
	LL res = 1;
	for(; y; y >>= 1)
	{
		if (y & 1) res = res * x % P;
		x = x * x % P;
	}
	return res;
}

inline void Euler()
{
	vis[1] = mu[1] = pk[1] = 1;
	for(re int i = 2; i <= N; i++)
	{
		if (!vis[i]) pr[++totp] = i, mu[i] = -1, pk[i] = fpow(i, k);
		for(re int j = 1; j <= totp && i * pr[j] <= N; j++)
		{
			vis[i * pr[j]] = 1, pk[i * pr[j]] = (LL)pk[i] * pk[pr[j]] % P;
			if (!(i % pr[j])) break;
			mu[i * pr[j]] = -mu[i];
		}
	}
	for(re int i = 1; i <= N / 2; i++) 
		f1[i] = ((LL)f1[i - 1] + (LL)pk[i] * i % P * mu[i] * mu[i]) % P,
		f2[i] = ((LL)f2[i - 1] + (LL)pk[i] * mu[i] + P) % P;
	for(re int i = 1; i <= N; i++) spk[i] = (pk[i] + spk[i - 1]) % P;
	for(re int i = 1; i <= N / 2; i++) f3[i] = ((LL)f3[i - 1] + 2LL * (spk[2 * i] - spk[i] + P) % P - pk[2 * i] % P + P) % P;
}

inline int query(int n)
{
	LL res = 0;
	for(re int l = 1, r; l <= n; l = r + 1)
	{
		r = n / (n / l);
		res = (res + (LL)(f2[r] - f2[l - 1] + P) % P * f3[n / l] % P) % P;
	}
	return res;
}

int main()
{
	scanf("%d%lld", &n, &k);
	Euler();
	LL ans = 0;
	for(re int l = 1, r; l <= n; l = r + 1)
	{
		r = n / (n / l);
		ans = (ans + (LL)(f1[r] - f1[l - 1] + P) % P * query(n / l)) % P;
	}
	printf("%lld
", ans);
}